I have a question about Hatcher's definition of $\Delta$-complex.
The standard n-simplex is defined as
$\Delta^n = \{(t_0, ..., t_n) \in \mathbb{R}^{n+1} | \sum_{i} t_i = 1, t_i \geq 0\}$
Let $\mathring{\Delta}^n$ denote the interior of a $\Delta$ n-simplex.
From p. 103 of Hatcher:
A $\Delta$-complex structure on a space X is defined as a collection of maps $\sigma_{\alpha} : \Delta^n \rightarrow X$, with n depending on the index $\alpha$, such that:
(i) The restriction $\sigma_{\alpha} | \mathring{\Delta}^n$ is injective, and each point of X is in the image of exactly one such restriction.
(ii) Each restriction of $\sigma_{\alpha}$ to a face of $\Delta^n$ is one of the maps $\sigma_{\beta} : \Delta^{n-1} \rightarrow X$.
(iii) A set $A \subset X$ is open iff $\sigma_{\alpha}^{-1}(A)$ is open in $\Delta^n$ for each $\sigma_{\alpha}$.
Question: Take the 1-dimensional simplex corresponding to the interval [0,1], with three maps (the 1-dimensional simplex map plus two 0-simplex maps as required by condition ii). Then this space fails to be a $\Delta$-simplex because the end points of the interval are not in the image of any interior point of either $\Delta^1$ (since they are endpoints) or $\Delta^0$ (since the interior of a point set is empty). Is this correct? It seems strange that a 1-simplex does not count itself as a $\Delta-$simplex under this definition.
Hatcher defines the interior of $\Delta^n$ as $\Delta^n\setminus\partial\Delta^n$. Here $\partial\Delta^n$ is defined as the union of the faces of $\Delta^n$, where a face can be obtained by removing one vertex and taking the convex hull of the remaining vertices. In the case $n=0$, there is only one vertex, so upon removing it the complex hull is empty and so $\partial\Delta^0$ is empty (actually, it is conventional to say that $\Delta^0$ has no faces rather than saying it has one face which is the empty set, but in any case $\partial\Delta^0$ is definitely the empty set). So the interior of $\Delta^0$ is not empty, but is instead $\Delta^0$ itself.